Simplify and expand the following expression: $ \dfrac{1}{r + 8}+ \dfrac{2}{3r - 12}+ \dfrac{4}{r^2 + 4r - 32} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the second term: $ \dfrac{2}{3r - 12} = \dfrac{2}{3(r - 4)}$ We can factor the quadratic in the third term: $ \dfrac{4}{r^2 + 4r - 32} = \dfrac{4}{(r + 8)(r - 4)}$ Now we have: $ \dfrac{1}{r + 8}+ \dfrac{2}{3(r - 4)}+ \dfrac{4}{(r + 8)(r - 4)} $ The least common multiple of the denominators is: $ (r + 8)(r - 4)$ In order to get the first term over $(r + 8)(r - 4)$ , multiply by $\dfrac{3(r - 4)}{3(r - 4)}$ $ \dfrac{1}{r + 8} \times \dfrac{3(r - 4)}{3(r - 4)} = \dfrac{3(r - 4)}{(r + 8)(r - 4)} $ In order to get the second term over $(r + 8)(r - 4)$ , multiply by $\dfrac{r + 8}{r + 8}$ $ \dfrac{2}{3(r - 4)} \times \dfrac{r + 8}{r + 8} = \dfrac{2(r + 8)}{(r + 8)(r - 4)} $ In order to get the third term over $(r + 8)(r - 4)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{4}{(r + 8)(r - 4)} \times \dfrac{3}{3} = \dfrac{12}{(r + 8)(r - 4)} $ Now we have: $ \dfrac{3(r - 4)}{(r + 8)(r - 4)} + \dfrac{2(r + 8)}{(r + 8)(r - 4)} + \dfrac{12}{(r + 8)(r - 4)} $ $ = \dfrac{ 3(r - 4) + 2(r + 8) + 12} {(r + 8)(r - 4)} $ Expand: $ = \dfrac{3r - 12 + 2r + 16 + 12}{3r^2 + 12r - 96} $ $ = \dfrac{5r + 16}{3r^2 + 12r - 96}$